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Lady_Fox [76]
2 years ago
9

10.Which device is designed to produce

Chemistry
2 answers:
Marta_Voda [28]2 years ago
5 0
The answer is a fan, it produces mechanical energy.

Please mark me as brainliest.
andreev551 [17]2 years ago
3 0

Which device is designed to produce mechanical energy?

Well, mechanical energy is the sum of potential and kinetic energy. Kinetic energy is a unit of energy in which is the reason why an object does movement, while potential energy is another unit of energy in which is the reason why an object doesn't move, or stops. While you can stop a fan or a lamp, it isn't automatic. It's manually. So the answer is (4) an oven because mechanical energy happens automatically, not manually, or for a greater reason.

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A 1.250-g sample of benzoic acid, C7H6O2, was placed in a combustion bomb. The bomb was filled with an excess of oxygen at high
Degger [83]

Answer:

3224 kJ/mol

Explanation:

The combustion of benzoic acid occurs as follows:

C₇H₆O₂ + 13/2O₂ → 7CO₂ + 3H₂O + dE

The change in temperature in the reaction is the change due the energy released, that is:

3.256K * (10.134kJ / K) = 33.00kJ are released when 1.250g reacts

To find the heat released per mole we have to find the moles of benzoic acid:

<em>Moles benzoic acid -Molar mass: 122.12g/mol-:</em>

1.250g * (1mol / 122.12g) = 0.0102 moles

<em />

The dE combustion per mole of benzoic acid is:

33.00kJ / 0.0102moles =

<em>3224 kJ/mol </em>

4 0
3 years ago
The atomic mass of an element is the sum of neutrons and A) atoms. B) electrons. C) isotopes. D) protons.
Elis [28]
Answer: D) protons.

The other option that would make the most sense would be electrons, however the mass of an electron is so small that is basically negligible, so it's not included in the atomic mass.
3 0
3 years ago
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10. (a) Describe how the structure of an alloy is different to a pure metal
Oxana [17]
1.For each of the following, give the name of an element from Period 4 (potassium to krypton), which matches the description.
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(1 Point)
an element that reacts with water to produce a lilac flame

2.For each of the following, give the name of an element from Period 4 (potassium to krypton), which matches the description.
Elements may be used once, more than once or not all.. Single line text.
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an element used as an inert atmosphere

3.For each of the following, give the name of an element from Period 4 (potassium to krypton), which matches the description.
Elements may be used once, more than once or not all.. Single line text.
(1 Point)
an element that has a valency of 3

4.Write a balanced chemical equation for the reaction between potassium and water. (Non-anonymous question). .
(1 Point)

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5.For each of the following, give the name of an element from Period 4 (potassium to krypton), which matches the description.
Elements may be used once, more than once or not all.. Single line text.
(1 Point)
an element with a fixed valency of 2 that not is not in group 2

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4 0
3 years ago
What are the four points of Dalton's Theory?<br>​
WARRIOR [948]

1. Elements are composed of atoms that are indestructible

2. All atoms of a given element are identical; same size/mass/chemical properties

3. Atoms of 1 element are different from the atoms of other elements

4. Compounds are composed of atoms with more than 1 element. The relative number of atoms for each element are of a given compound are always going to be the same.

(Extra one) 5. Chemical reactions are only ever going involve the rearrangement of the atoms. Atoms are not created/destroyed during the chemical reactions. (Law of Conservation of Mass: nothing can ever be created or destroyed.)

3 0
3 years ago
How many grams of sodium phosphate monobasic would we add to a liter and how many grams of sodium phosphate dibasic would we add
Nostrana [21]

Answer :

The correct answer   for Mass of Na₂HPO₄ = 4.457 g and mass of  NaH₂PO₄  = 8.23 g

Given :  pH = 6.86

Total concentration of Phosphate buffer = 0.1 M

Asked : Mass of  Sodium phosphate monobasic (NaH₂PO₄) = ?

Mass of  Sodium phosphate dibasic(Na₂HPO₄)= ?

Following steps can be done to find the masses of NaH₂PO₄ and Na₂HPO₄ :

(In phosphate buffer , Na+ ion from  NaH₂PO₄ and Na₂HPO₄ acts as spectator ion , so only H₂PO₄⁻ and HPO₄²⁻ will be considered )

<u>Step 1 : To find pka </u>

H₂PO₄⁻  <=> HPO₄²⁻  

The above reaction has pka = 7.2 ( from image shown )

<u>Step 2 : Plug values in Hasselbalch- Henderson equation </u>.

Hasselbalch -Henderson equation is to find pH  for buffer solution which is as follows :

pH = pka + log\frac{[A^-]}{[HA]}

pH = 6.86         pKa = 7.2

6.86 = 7.2 + log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}

Subtracting  both side by 7.2

6.86-7.2 = 7.2 -7.2+ log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}

-0.34 =  log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}

Removing log

10^-^0^.^3^4 =   \frac{[HPO_4^2^-]}{[H_2PO_4^-]}

\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457 ---------------- equation (1)

<u>Step 3 : To find  molarity of H₂PO₄⁻ and HPO₄²⁻</u>

Total concentration of buffer = [H₂PO₄⁻] + [HPO₄²⁻] = 0.1 M

Hence,  [H₂PO₄⁻ ] + [ HPO₄²⁻ ] =  0.1 M

Assume [H₂PO₄⁻ ] = x

So ,  [x ] + [ HPO₄²⁻ ] =  0.1 M

[ HPO₄²⁻ ] =  0.1 - x

Step 4 : Plugging value of [H₂PO₄⁻ ]  and  [ HPO₄²⁻ ]

[H₂PO₄⁻ ]  = x

 [ HPO₄²⁻ ] = 0.1 - x

Equation (1) = >\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457

Plug value of [H₂PO₄⁻ ]  and  [ HPO₄²⁻ ] ( from step 3 ) into equation (1)  as :

\frac{[0.1 - x ]}{[ x]} = 0.457

Cross multiplying

0.1 - x  = 0.457 x

Adding x on both side

0.1 -x + x = 0.457 x + x

0.1  = 1.457 x

Dividing both side by 1.457

\frac{0.1}{1.457} = \frac{1.457 x }{1.457}

x = 0.0686 M

Hence , [H₂PO₄⁻ ]  = x  = 0.0686 M

 [ HPO₄²⁻ ] = 0.1 - x

 [ HPO₄²⁻ ]  =   0.1 - 0.0686  

[ HPO₄²⁻ ] = 0.0314 M

Step 5 : To find moles of  H₂PO₄⁻ ( NaH₂PO₄) and HPO₄²⁻ (Na₂HPO₄ ) .

Molarity is defined as mole of solute per 1 L volume of solution .

Molarity of NaH₂PO₄ = 0.0686 M  or 0.0686 mole per 1 L

Molarity of Na₂HPO₄ = 0.0314 M  or 0.0314 mole per 1 L

Since  that volume of buffer solution  is 1 L , so Molarity  = mole

Hence Mole of NaH₂PO₄  = 0.0686 mol

Mole of Na₂HPO₄ = 0.0314 mol

<u>Step 6 : To find mass  of Na₂HPO₄  and NaH₂PO₄ </u>

Moles of  Na₂HPO₄  and NaH₂PO₄  can be converted to their masses using molar mass as follows :

Molar mass of  Na₂HPO₄  = 141.96 \frac{g}{mol}

Molar mass of NaH₂PO₄ = 119.98 \frac{g}{mol}

Mass (g) = mole (mol)* molar mass(\frac{g}{mol})

Mass of Na_2HPO_4 = 0.0314 mol * 141.96 \frac{g}{mol}

Mass of Na₂HPO₄ = 4.457 g

Mass of NaH_2PO_4 = 0.0686 mol * 119.98 \frac{g}{mol}

Mass of  NaH₂PO₄  = 8.23 g

5 0
3 years ago
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